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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>From (<a href="" class="xref" data-knowl="./knowl/eq5_6.html" title="Equation 5.3.3">(5.3.3)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
y_2(x)=\sum_{n=0}^{\infty} a_n x^n,\quad \textrm{with}~a_0=0, a_1=1.
\end{equation*}
</div>
<p class="continuation">In (<a href="" class="xref" data-knowl="./knowl/eq5_7.html" title="Equation 5.3.4">(5.3.4)</a>), let <span class="process-math">\(a_0=0, a_1=1\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
n=0:&amp;\quad a_2=0,\\
n=1: &amp;\quad a_3=-\frac{(\alpha-1)(\alpha+2)}{3!},\quad (k=1)\\
n=2: &amp;\quad a_4=0,\\
n=3: &amp;\quad a_5=-\frac{(\alpha-3)(\alpha+4)}{5 \cdot 4}a_3=\frac{(\alpha-1)(\alpha-3)(\alpha+2)(\alpha+4)}{5!}, \quad (k=2)\\
n=4: &amp;\quad a_6=0,\\
n=5: &amp; \quad a_7=-\frac{(\alpha-5)(\alpha+6)}{7\cdot 6}=-\frac{(\alpha-1)(\alpha-3)(\alpha-5)(\alpha+2)(\alpha+4)(\alpha+6)}{7!},\quad (k=3)\\
&amp;\quad \cdots\\
&amp;\quad a_{2k}=0,\\
&amp;\quad a_{2k+1}=(-1)^k \frac{(\alpha-1)(\alpha-3)(\alpha-5)\cdots(\alpha-2k+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2k)}{(2k+1)!}, \quad k\geq1.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
y_2=\sum_{n=0}^{\infty} a_n x^n=x+\sum_{n=2}^{\infty} a_n x^n=&amp;x+\sum_{k=1}^{\infty} a_{2k} x^{2k}+\sum_{k=1}^{\infty} a_{2k+1} x^{2k+1}\\
&amp;=x+\sum_{k=1}^{\infty} (-1)^k \frac{(\alpha-1)(\alpha-3)\cdots(\alpha-2k+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2k)}{(2 k+1)!} x^{2k+1}.
\end{aligned}
\end{equation*}
</div>
<span class="incontext"><a href="sec5_3.html#p-223" class="internal">in-context</a></span>
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